Sunday, April 7, 2013

Work Problems in Mathematics

Tunay na malaking challenge o problema ang kinakaharap ng mga mag-aaral sa pagsagot ng mga word problems sa Math kapag ito ay isang Work Problem.



Ang Work Problem ay isang word problem kung saan dalawa o higit pang tao ang nasasangkot na gumagawa ng isang trabaho na may IBA'T IBANG bilis (rate or speed) sa paggawa.

Halimbawa, kayang tapusin ni Juan ang pagbubunot ng sahig sa loob lamang ng tatlumpung minuto. Dalawampung minuto naman kapag si Pedro ang gumawa. Kung sabay nilang bubunutin ang sahig, ilang minuto nila itong tatapusin?

PARAAN:

1) Isulat natin ang ating mga Given at Variables:
 
    Juan = 30 min.   ===> ang oras (time) bahagi ng trabahong kanyang gagawin = 1/30

    Pedro = 20 min. ===> ang oras (time) bahagi ng trabahong kanyang gagawin = 1/20

   Kabuuan trabaho = x  ===>    ang kabuuang  oras ng trabahong gagawin nina Juan at Pedro = 1/x

TANDAAN:  

Mapapansin natin na ang RECIPROCAL ng numero ang bahagi ng trabahong gagawin.

Ano ba ang reciprocal? Ito ay ang kabaliktaran ng isang bilang.

Halimbawa, ang reciprocal ng 2 ay 1/2. Kaya ang reciprocal ng 1/2 ay 2.
Ang reciprocal ng  2/3 ay 3/2.
Ang reciprocal ng x ay 1/x.
Kuha na?

 2) Isulat ang EQUATION

1/30 + 1/20 = 1/x

3) SAGUTIN

Hanapin ang LCD o Less Common Denominator==> ang pinakamaliit na numerong maaaring hatiin sa dalawang denominator na 30 at 20. Kung nahihirapan kayong hanapin ang LCD, maaaring i-multiply ninyo ang dalawang  denominators na 30 at 20 at ito ang kuhanin. Maaari ring hanapin ang LCM o Less Common Multiple ==> ang pinakamaliit na product ng 2 denominators kung i-mumultiply sila sa counting numbers ( 1, 2, 3, 4 ... and so on)

Kung gayon ===>  30 x 20 = 600

Pagkatapos, i-multiply ang nakuhang product sa itaas sa BUONG EQUATION.

600 x (1/30 + 1/20 = 1/x)

20 + 30 = 600/x

50 = 600/x

x = 600/50

x  = 12 min

Sa pagkuha ng LCD, maaaring isama na rin ang denominator ng ikatlong term o yong nasa kaliwa ( 1/x).

Kaya ang makukuhang product ay 600x
Kung magkagayun,

600x(1/30 + 1/20 = 1/x)

20x + 30x = 600

50x = 600

x = 600/50

x = 12 min.

SUMMARY

Sa kabuuan, ang formula sa pagsagot ng work problem kung saan iba't iba ang bilis ng paggawa ay:

1/t1 + 1/t2 = 1/tb

kung saan, t1 = time o oras ng unang tao
                 t2 = time o oras ng ikalawang tao
                 tb = kabuuang oras (time) ng dalawa

Ang pormulang nabanggit sa itaas ay maaaring gamitin sa dalawa o higit pang tao at sa mga pipes filling up a tank problem.

Halimbawa (from http://www.onlinemathlearning.com/math-work-problems.html)


Peter can mow the lawn in 40 minutes and John can mow the lawn in 60 minutes. How long will it take for them to mow the lawn together?
Solution:

Step 1: Assign variables:
Let = time to mow lawn together
Step 2: Use the formula:
1/40+1/60=1/x
Step 3: Solve the equation
The LCM of 40 and 60 is 120
 Multiply both sides with 120
solve the eqn
Answer: The time taken for both of them to mow the lawn together is 24 minutes.

Work Problem with one unknown time


Paano naman kung ang isa sa dalawa (o higit pang tao) ay hindi alam ang oras ng paggawa?

Halimbawa:

Jack can paint a house in 6 hours. If his brother, Dave helps, it takes them four and two-thirds hours. How long will it take Dave working alone?

1) Isulat ang Given at Variable

  Jack = 6 hours   : Time =  1/6 (Reciprocal)

 Dave  = x   : Time = 1/x  (Reciprocal)

 Total Time =   4 2/3   = 14/3  :  Time = 3/14  (Reciprocal)

 2) Isulat ang Equation or Formula

   1/6 + 1/x = 3/14

 3) Sagutin ang Equation

Kung ayaw hanapin ang LCD, i-multiply na lang ang lahat ng denominators ( 6, x at 14)

6 times x times 14 = 84x

I-multiply ang nakuhang produkto sa itaas (84x) sa Equation

Kaya,

84x ( 1/6 + 1/x = 3/14)

84x/6 + 84x/x = 84x (3/14)

14x + 84 = 18x  

18x = 14x + 84

4x = 84

x = 21 hours

Checking:

1/6 + 1/x = 3/14
since x = 21

1/6 + 1/21 = 3/14

6x21x14 = 1764

1764(1/6 + 1/21 = 3/14)

1764/6 + 1764/21 = 1764 (3/14)

294 + 84 = 5292/14

378 = 378

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Friday, March 15, 2013

Distance Word Problem

Isa sa mahirap na parte ng pag-aaral sa Math ay ang pag-solve sa mga word problem dahil ito ay nangangailangan ng pag-aalisa o masusing paggamit ng common sense at imahinasyon. Ang mahirap na bahagi ng pagsagot sa mga word problem ay ang pag-setup ng equation. Para sa araling ito, ang ating pagtutuunan ng pansin ay ang pagsagot sa Distance Problem.



Ang "distance word problems" ay tinatawag din "uniform rate problems" kung saan ang isang bagay ay naglalakbay ng iisang bilis lamang o paiba-iba ang bilis sa isang paglalakbay. Kalimitan, ang isang distance problem ay sumasagot sa tanong na 1) kung gaano kabilis ang takbo; 2) gaano kalayo; o 3) gaano katagal.

Distance Formula

Ang distance formula ay  d = rt,

kung saan ang d ay ang distansya; r ay ang uniform o average speed at ang t ay ang oras.


Babala sa paggamit ng distance formula.

Siguraduhin lamang na ang unit para sa oras (t) at distansya (d) ay PAREHO sa unit ng rate of speed (r).
Halimbawa kung ang rate (r) ay kilometer per hour, dapat ang distance ay kilometer at ang oras (time) ay hour.

Halimbawa ng Distance Problem:

My taxi is travelling at two different speeds from my house to the airport.On asphalt road, it travels at 105 mph. On concrete road, it travels at 115 mph. I left at 1:00 PM and arrived at the airport at 6:00 PM. The distance of my house from the airport is 555 miles. For how long did my taxi travel from each speed?

Solution:

Istratehiya sa pagsagot ng word problem:

1) Isaulo at ibigay ang general formula ng distance"

distance = rate of speed x time or d = rt

2) Dahil may dalawang speed ang taxi, ibigay ang bawat formula o equation nito.
a) Sa aspaltadong daan, ang speed o bilis  (r) ng taxi ay 105 mph (miles per hour).
Dahil d = rt, ang equation  nito ay
d = 105t
     
b) Sa sementadong daan, ang speed (r) ng taxi ay  115 mph.
Ang equation nito ay  555 - d = 115 ( 5-t).

Busisiin natin ang pangalawang equation.

Bakit ang distance (d) ay naging 555-d?

Ang total na distansya ng airport at bahay ay 555 miles (given). Dahil ang distance na itinakbo ng taxi sa aspaltadong daan ay d, ang matitirang distansyang tatakbuhin ng taxi sa sementadong daan ay 555 - d na lamang.

Saan kinuha ang 115(5-t)?

Ang 115 ay ang speed ng taxi sa sementadong daan. Bakit naging 5-t and oras ng taxi? Ang kabuuang oras ng paglalakbay ay 5 oras ( 6PM - 1PM = 5 hours). Dahil ang oras sa aspaltadong daan ay t na, natural lamang na ibawas ang t sa total na oras kaya naging 5 - t.

3) Pagsamahin o i-ADD ang dalawang equation.

Equation 1 :  d = 105t
Equation 2:    555 - d = 115(5-t)

d + 555 - d = 105t + 115(5-t)

555 = 105t + 575 - 115t

555 = 575 - 10t

-20 = -10t

t = 2 hours

Ang t = 2 hours ay ang sagot sa unang equation na ang ibig sabihin ay 2 oras ang tinakbo ng taxi sa aspaltadong daan.

Ilang oras naman ang tinakbo ng taxi sa sementadong daan?

5 - t   o   5 - 2 = 3 hours

Tatlong oras ang tinakbo ng taxi sa sementadong daan.

Hintayin ang ilang halimbawa para sa pagsagot ng distance problem.

Examples from  Purple Math

  • A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? How far did the boat travel one way?
       
    • drt
      downstreamdb + 33
      upstreamdb – 34
      total2d---7

    (It may turn out that I won't need the "total" row.)
    I have used "b" to indicate the boat's speed. Why are the rates "b + 3" and "b – 3"? Because I actually have two speeds combined into one on each trip. The boat has a certain speed (the "speed in calm water" that I'm looking for; this is the speed that registers on the speedometer), and the water has a certain speed (this is the "current"). When the boat is going with the current, the water's speed is added to the boat's speed. This makes sense, if you think about it: even if you cut the engine, the boat would still be moving, because the water would be carrying it downstream. When the boat is going against the current, the water's speed is subtracted from the boat's speed. This makes sense, too: if the water is going fast enough, the boat will still be going downstream (a "negative" speed, because the boat would be going backwards at this point), because the water is more powerful than the boat. (Think of a boat in a cartoon heading toward a waterfall. The guy paddles like crazy, but he still goes over the edge.)

      drt
      downstreamd = 3(b + 3)b + 33
      upstreamd = 4(b – 3)b – 34
      total2d---7
    Using "d = rt", the first row (of the revised table above) gives me:

      d = 3(b + 3)

    The second row gives me:

      d = 4(b – 3)
    Since these distances are the same, I will set them equal:

      3(b + 3) = 4(b – 3)

    Solve for b. Then back-solve for d.
In this case, I didn't need the "total" row.
  • With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.
       
    • drt
      tailwind1120p + w7
      headwind1120p – w8
      total2240---15
    (I probably won't need the "total" row.) Just as with the last problem, I am really dealing with two rates together: the plane's speedometer reading, and the wind speed. When the plane turns around, the wind is no longer pushing the plane to go faster, but is instead pushing against the plane to slow it down.  Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
    The first row gives me:
      1120 = 7(p + w)

    The second row gives me:
      1120 = 8(p – w)

    The temptation is to just set these equal, like I did with the last problem, but that just gives me:
      7(p + w) = 8(p – w)

    ...which doesn't help much. I need to get rid of one of the variables.
    I'll take that first equation:
      1120 = 7(p + w)

    ...and divide through by 7:
      160 = p + w
    Then, subtracting from either side, I get that p = 160 – w. I'll substitute "160 – w" for "p" in the second equation:
      1120 = 8([160 – w] – w)
      1120 = 8(160 – 2w)
    ...and solve for w. Then I'll back-solve to find p.
  • A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per second and through steel at 16,500 feet per second. How far away is that spike? (Round to one decimal place.)
      
    • drt
      aird = 1100t1100t
      steel= 16,500(t – 6)16,500t – 6
      total------6
    However long the sound took to travel through the air, it took six seconds less to propagate through the steel. (Since the speed through the steel is faster, then that travel-time must be shorter.) I multiply the rate by the time to get the values for the distance column. (Once again, I didn't need the "total" row.)
    Since the distances are the same, I set the distance expressions equal to get:

      1100= 16,500(t – 6)
    Solve for the time t, and then back-solve for the distance d by plugging t into either expression for the distance d.
Sana ay naunawaan.